 Absorbing state

For a finite, time-homogeneous Discrete-time Markov chain with row-stochastic transition matrix $P$, a state $a$ is absorbing exactly when $P_{aa}=1$. Nonnegativity and the row sum then force $P_{aj}=0$ for $j\ne a$: once entered, $a$ is never left. Equivalently, $\{a\}$ is a closed communicating class. A larger closed communicating class may keep moving internally; closure means confinement to a set, whereas absorption at a state means remaining at one point.

An absorbing state is useful for recording a terminal outcome: after a toy routine finishes, keep its outcome in the state space so that the same transition matrix still describes subsequent steps. Calling a finite Markov chain absorbing usually also requires that some absorbing state be accessible from every state. Merely containing an absorbing state is insufficient: another closed communicating class could prevent absorption. See Grinstead and Snell, Introduction to Probability, section 11.2 for this convention.

Consider a constructed routine with unfinished modes $u,v$ and terminal outcomes $A,B$. All the following transition probabilities are chosen for illustration, not estimated from observations. In state order $(u,v,A,B)$, takeThe communicating class $\{u,v\}$ is open, while $\{A\}$ and $\{B\}$ are closed communicating classes consisting of absorbing states. Both terminal outcomes are accessible from both unfinished modes, sometimes by an indirect path.

Let $T=\inf \{n\ge 0:X_n\in \{A,B\}\}$ be the terminal hitting time. Write $h_i={\mathbb{P}}_i(X_T=A)$ for the absorption probability and $t_i={\mathbb{E}}_{i}T$ for the expected hitting time, with the subscript denoting the starting state. First-step analysis givesHere $h_A=1$, $h_B=0$, and $t_A=t_B=0$. Solving these two pairs of linear equations yieldsStarting at $u$ therefore ends at $A$ with probability $0.6$ and takes $3.2$ steps on average to reach either terminal state. The terminal hitting time counts the transition into the outcome, but no later self-transitions. Outcome and duration answer different questions: from $v$, outcome $A$ is less likely, yet termination is faster on average.

The matrix power $P^2$ gives the probability distribution after two steps from $u$ asThus ${\mathbb{P}}_u(T>2)=1/2$, and the probability of having already reached $A$ is $3/8$, still below its eventual absorption probability $3/5$. From either unfinished mode, each next step has terminal probability at least $1/4$. Consequently ${\mathbb{P}}_i(T>n)\le (3/4{)}^n$, proving termination almost surely and finite expected hitting time directly. The Fundamental matrix of a finite absorbing Markov chain packages both calculations and counts visits to unfinished modes.